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A particle of mass m is moving in YZ-plane with a uniform velocity `v` with its trajectory running parallel to `+ve` Y-axis and intersecting Z-axis at `z=a` in figure. The change in its angular momentum about the origin as it bounces elastically form a wall at y=constant is |
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Answer» Correct Answer - b The initial velocity is `vecv_(i) = vhatey` After reflection from the wall, the final velocity is `vecv_(f) = -vhatey` The trajectory is given as `vecr= yhate_(y) + ahate_(z)` Hence, the change in angular momentum is `DeltavecL=vecr xx m(vecv_(f) - vecv_(i))` `=(yhate_(y) + avece_(z)) xx (-2mhatey)` `=2"mva"vece_(x) [therefore hate_(y) xx hate_(y) = 0` and `hate_(z) xx hate_(y) = -hate_(x)]` |
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